Question

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and accelerates at 15.0 km/h2 for 20.0 min. Car C starts out traveling at 32.0 km/h and accelerates at 40.0 km/h2 for 30.0 min. Car D starts out traveling at 110.0 km/h and decelerates at 60.0 km/h2 for 45.0 min. Which car's final speed was closest to 15 m/s? A. A B. B C. C D. D

Answer 1

1 kilometre=1000 metre

      1 hour = 3600 second

       $1\ km/hr=\frac{1000}{3600} m/s$

       $1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

                                         $35.0\ km/hr=35*\frac{5}{18} m/s$

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  $25\ km/hr^2$

                                            $=\ 25*\frac{1000}{3600*3600} m/s^2$

                                            $=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    $15\ km/hr^2$

                                    $=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    $40\ km/hr^2$          

                                                            $=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  $60\ km/hr^2$

                                                                      $=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.